工程材料科学与设计 答案

Problems - Chapter 5 1. FIND: Calculate the stress on a tensioned fiber.

GIVEN: The fiber diameter is 25 micrometers. The elongational load is 25 g. ASSUMPTIONS: The engineering stress is requested.

DATA: Acceleration due to gravity is 9.8 m/sec 2. A Newton is a kg-m/sec 2. A Pascal is a N/m 2. A MPa is 106 Pa.

SOLUTION: Stress is force per unit area. The cross-sectional area is πR 2 = 1963.5 square micrometers. The force is 25 g (kg/1000g)(9.8 m/sec 2) = 0.245 N. Thus, the stress is

σ = F/A =

024519601001252

6

2

..N um

um m MPa

⨯☞☟☝✋ ☺=

COMMENTS: You must learn to do these sorts of problems, including the conversions. 2. GIVEN: FCC Cu with a o = 0.362nm

REQUIRED: A) Lowest energy Burgers vector, B) Length in terms of radius of Cu atom, C) Family of planes

SOLUTION: We note that the Burgers vector is the shortest vector that connects crystallographically equivalent positions. A diagram of the structure is shown below:

FCC structure with (111) shown

We note that atoms lying along face diagonals touch and are crystallographically equivalent. Therefore, the shortest vector connecting equivalent positions is ½ face diagonal. For example,

one such vector is

10]

1[ 2

a o as shown in (111).

A. The length of this vector is

0.256nm = 2

0.362

= 2

a

= 4

a

+ 4

a o 2

o 2

o

B. By inspection, the size of the vector is 2 Cu atom radii.

C. Slip occurs in the most densely packed plane which is of the type {111}. These are the

smoothest planes and contain the smallest Burgers vector. This means that the dislocations move easily and the energy is low.

3. GIVEN:∣b∣ = 0.288nm in Ag

REQUIRED: Find lattice parameter

SOLUTION: Recall the Ag is FCC. For FCC structures the Burgers vector is ½ a face

diagonal as shown. We see that

4. A. FCC structure

The (111) plane is shown in a unit cell with all atoms shown. Atoms touch along face diagonals. The (111) plane is the most closely packed, and the vectors shown connect equivalent atomic position. Thus 10]1[ 2

1

= b etc. Then in general >110< 2

a

= b

B. For NaC1

We see that the shortest vector connecting equivalent positions is 10]1[ 2a

as shown. This

direction lies in both the {100} and {110} planes and both are possible slip planes. However {110} are the planes most frequently observed as the slip planes. This is because repulsive interionic forces are minimized on these planes during dislocation motion. Thus we expect 1/2<110> Burgers vectors and {110} slip planes.

5.

GIVEN:

Mo crystal

0.272nm = b

a o = 0.314nm

REQUIRED: Determine the crystal structure.